Problem: $f(x)=\dfrac{1}{3+x^2}$ Find a power series for $f$. Choose 1 answer: Choose 1 answer: (Choice A) A $ \frac{1}{3}-\frac{1}{9}x^2+\frac{1}{27}x^4+\ldots +\frac{1}{3}\left(-\frac{1}{3}\right)^n\left( x \right)^{2n}+\ldots$ (Choice B) B $ 1-\frac{1}{3}x^2+\frac{1}{9}x^4+\ldots +\left(-\frac{1}{3}\right)^n(x )^{2n}+\ldots$ (Choice C) C $ 1+x^2+x^4+\ldots +x ^{2n}+\ldots$ (Choice D) D $ 1-x^2+x^4+\ldots +(-1)^nx ^{2n}+\ldots$
Answer: We have to factor a $3$ out of the denominator in order to rewrite the expression as a geometric series. $ f(x)= \frac{1}{3+x^2}=\frac{1}{3(1+\frac{1}{3}x^2)}=\frac{1}{3}\cdot\frac{1}{1+\frac{1}{3}x^2}$ We now have a geometric series with first term $a\text{ }=\text{ }1$ and common ratio $r\text{ }=\text{ }-\frac{1}{3}x^2$, with a constant multiplier of $\frac{1}{3}$ in front. Therefore, the series is as follows. $\frac{1}{3}\Big[1-\frac{1}{3}x^2+\frac{1}{9}x^4+\ldots +\left(-\frac{1}{3}\right)^n x ^{2n}+\ldots \Big]$ $ = \frac{1}{3}-\frac{1}{9}x^2+\frac{1}{27}x^4+\ldots +\frac{1}{3}\left(-\frac{1}{3}\right)^n x ^{2n}+\ldots $